Problem: In the right triangle shown, $AC = 4$ and $BC = 6$. What is $AB$ ? $A$ $C$ $B$ $4$ $6$ $?$
Explanation: We know $a^2 + b^2 = c^2$ We want to find $c$ ; let $a = 6$ and $b = 4$ So $c^2 = 6^2 + 4^2 = 52$ Then, $c = \sqrt{52}$ Simplifying the radical gives $c = 2\sqrt{13}.$